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Find the volume of the solid formed when the area bound by the curve y=4-x^2 and the x axis is rotated one complete revolution about the y axis

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first find the critical points and use integration
or just consider one side of it
can you just tell me the answer , because i am getting the wrong answer. I got 16 pi units but the answer is 8 pi units and i don't know where i went wrong.

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Other answers:

*units ^3
please dont ask for "just an answer"
|dw:1342094312271:dw| this can be done by shells or disks; which way are you most comfortable with?
i think i know why youre twice as big as the answer; your prolly integrating the whole thing across
your in effect sweeping the same volume twice when you do that
the answer is 8pi
|dw:1342094460320:dw| if we do shells:\[\int_{0}^{2}2pix(4-x^2)dx\] \[2pi\int_{0}^{2}x(4-x^2)dx\] \[2pi\int_{0}^{2}4x-x^3\ dx\] \[2pi\left(\frac{1}{2}4x^2-\frac{1}{4}x^4\right)_{0}^{2}\]
oh i think i understand. I apologise amistre64 if i sounded rude before. I had no intentions whatsoever to offend you.
no offense taken :)
@amistre64 i thought it was around y axis, shouldn't be dy not dx ?
depends on your method; i used the shell method, which opens up along the x axis the disc method would travel along the y
teh disc method we would have to convert y=x into its inverse x=y, which isnt that hard to do, but just takes extra steps :)
oh i see, so for the shell method where does the x come from in x(4-x^2)
think of the shell method as finding the area of a sheet of paper, or a flattened out tin can. the width of the paper is the circumference of the base circle|dw:1342095061762:dw| the height of the paper is the function that it hits along the way. Area = base (2pi x) * height (f(x))
OH I GET IT NOW ! thanks @amistre64
youre welcome

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